The Finite Square Well G.R.Dixon INTRODUCTION. This monograph discusses the quantum theory of a finite square well. It employs the spinning spiral model discussed in a previous article. A brief review of that model is provided in Sect. 1. The spinning spiral model is analogous to a circularly polarized electromagnetic wave when such a wave propagates in the +x-direction. At any given x the electromagnetic field vectors can be drawn in a real plane oriented parallel to the yz-plane. The quantum wave function (Y) is complex, however, and must be drawn in a complex plane. With no loss of rigor such a plane can also be visualized parallel to the yz-plane, centered on the x-axis with (a) its real axis parallel to the z-axis, and (b) its imaginary axis parallel to the y-axis. 1. REVIEW. Zero Potential Energy. When an ensemble of monoenergetic particles moves in the positive or negative x-direction, a rotating spiral provides a convenient model for visualizing the quantum wave function in space and time. A left-handed spiral, rotating clockwise (looking in the positive x-direction), can represent the wave function for particles moving in the positive x-direction. Similarly, a right-handed spiral (also rotating clockwise) can represent the wave function for particles moving in the negative x-direction. The threads of the left-handed spiral move in the positive x-direction as the spiral spins; the threads of the right-handed spiral move in the negative x-direction. At any given x the wave function for either particle ensemble can be visualized as a vector in a complex plane oriented orthogonal to the x-axis and with its origin on the x-axis. The vector’s tail would then coincide with the x-axis, and its head would be attached to the associated spiral. As previously mentioned, with no loss of rigor the complex plane’s real axis can be aligned parallel to the z-axis, and its imaginary axis can be aligned parallel to the y-axis. The formulas for the wave function of particles traveling toward positive (Y(+)) and negative (Y(-)) x are: , (1.1a) . (1.1b) Note that at times t other than zero, Y(+)(x,t) or Y(-)(x,t) is simply Y(+)(x,0) or Y(-)(x,0) rotated through an angle -wt in the complex plane. For simplicity we shall consider things only at times t=0, in which case Eqs. 1.1a and b can be simplified to , (1.1c) . (1.1d) The real constant k (the wave number) is related to either spiral’s wavelength (l) by: . (1.2a) And the real constant w is related to either spiral’s spin frequency (n) by: . (1.2b) DeBroglie suggested that the wavelength is related to particle momentum magnitude by . (1.3a) He also suggested that the spiral’s spin frequency is related to the energy (which equals the particle kinetic energy when the potential energy, U, is zero): . (1.3b) ln, the spiral thread speed, is then half the particle speed: . (1.4) Born suggested that the squared magnitude of the wave function (which is always a real number) is the probability density for finding a particle at a given location. For example, the probability of finding a particle traveling in the positive x-direction and in the interval x to x+dx (or (x,dx) for short) is: . (1.5) Note that P in this simple example is actually independent of x (and t). That is, the magnitude of Y(+/-) remains constant as x varies. Only the real and imaginary parts are functions of x. Normalization … the requirement that the integral of Pdx over all x be equal to unity (i.e., the requirement that the particle will be found somwhere) … indicates that |Y(+)(0)| and/or |Y(-)(0)| in Eqs. 1.1c and d are infinitesimal. The important point is that (as classically expected), given an ensemble of monoenergetic particles traveling in the positive x-direction, Pdx is the same at all x. The same holds for an ensemble of particles traveling in the negative x-direction. The first "quantum surprise" occurs when two such ensembles are combined. For in this case the probability of finding a particle traveling in either direction, in some interval (x,dx), is . (1.6) With no loss of rigor we can let Y(-)(0)=-Y(+)(0) in Eqs. 1.1c and d. Having done this and invoking the identity eiq=cos(q) + i sin(q), we find on combining Eqs. 1.1c and d that . (1.7) It is noteworthy that Y(x) is a sinusoid … its magnitude varies with x. Thus P, the probability density, goes as sine squared: . (1.8) This consequence of adding Y(+)(x) and Y(-)(x) is certainly different from the classical expectation that P would be the same at all x (and t). Infinite Barriers. Fig. 1.1 depicts an infinite potential energy barrier at x=0. U is zero for all x<0, and U is infinite for all x>0. Both classically and in quantum theory, particles traveling along the negative x-axis and in the positive x-direction are turned back at x=0. Figure 1.1 An Infinite Potential Energy Barrier Since all particles are turned back at x=0, P(x>0)=0. And of course P(x<0) is proportional to sine squared and is given by Eq. 1.8. It is logically plausible that P must be continuous in x. This being the case, we can again specify with no loss of rigor that , (1.9a) . (1.9b)   Square Wells. Fig. 1.2 depicts an infinite square well. Each wall constitutes an infinite barrier to particles trapped in the well. Particles traveling in the positive x-direction are turned back at x=L; those traveling in the negative x-direction are turned back at x=0. Assuming the particles are monoenergetic, Eqs. 1.1c and d again specify the wave functions for the two ensembles. And of course Eq. 1.7 gives their sum. Figure 1.2 An Infinite Square Potential Energy Well The requirement that P be continuous in x can be satisfied at x=0 by again stipulating that Y(+)(0) and Y(-)(0) are given by Eqs. 1.9a and b. At the instant t=0 the head of Y(+)(x) traces along a left-handed spiral to the wall at x=L. Similarly Y(-)(x) traces along a right-handed spiral. The requirement that Y(+)(L) + Y(-)(L) also equal zero produces the second "quantum surprise": only discrete half wavelengths can be accommodated by the infinite well: (1.10) According to deBroglie, this means that only discrete momentum magnitudes (and thus only discrete kinetic energies) can be accommodated by an infinite well. For example, the lowest energy (or "ground") state must have a wavelength of l/2: . (1.11) But according to deBroglie, . (1.12) Thus the ground state energy, Eo, is: . (1.13) Similarly, , (1.14) In general: . (1.15) Let us trace a complete circuit from x=0 to x=L and back again. We shall consider the ground state, where L equals a half wavelength. By the time we have moved along the left-handed spiral for Y(+)(x), from x=0 to x=L, Y(+)(x) has transitioned from Yo to -Yo At x=L, Y(+)(L) is reflected and becomes Y(-)(L), whose value at x=L is Yo. Thus Y(+)(L) experiences a phase change into Y(-)(L) upon reflection. The spiral for Y(-)(x) is right-handed. In returning from x=L to x=0, Y(-)(x) goes from Yo to -Yo. It too has a phase change at x=0, becoming Y(+)(0), and the cycle repeats. Any wavelength not satisfying Eq. 1.10 will not produce unique values of Y(+)(x) and/or Y(-)(x) between the two walls. 2.  A STRATEGY FOR SOLVING THE FINITE WELL. Fig. 2.1 depicts a finite potential energy well. We shall attempt to solve for Y(+)(x) and Y(-)(x) (and thus for Y(x)= Y(+)(x)+Y(-)(x)) inside the well and outside of it where x<0. (The symmetric solutions would apply at x>L.) Figure 2.1 A Finite Square Well For particles trapped in the well and traveling in the negative x-direction, the wall at x=0 constitutes a finite barrier. Similarly for particles traveling in the positive x-direction when they encounter the wall at x=L. It thus behooves us to begin our analysis of the finite well by considering the wave function for particles that encounter a finite barrier. 3.  THE FINITE BARRIER. Fig. 3.1 depicts a finite potential barrier at x=0. Monoenergetic particles, with kinetic energy mv2/20, , (3.2a) . (3.2b) Summing Eqs. 3.2a and b, we find that (3.3) . And , (3.4) . (3.5) Or, since sin(2q)=2 sin(q)cos(q), . (3.6) Specifically, at x=0: , (3.7a) . (3.7b) In considering the situation at negative x, we might begin by assuming that E, the total energy, is a system constant: . (3.8) Since U(x>0)=0, the total energy equates to mvx2/2 at positive x. But for x<0 Uo is greater than E. Eq. 3.8 therefore suggests the non-classical result that the kinetic energy is negative at x<0. Using primes to indicate quantities in the "classically forbidden zone," the requirement that the kinetic energy be negative can mathematically be satisfied by making vx’ imaginary, say: , (3.9a) . (3.9b) By deBroglie, the wave numbers for Y’(+) and Y’(-) must then also be imaginary: , (3.10a) . (3.10b) Substituting in Eqs. 3.2a and b produces: , (3.11a) . (3.11b) Or, defining a to be: , (3.12) Eqs. 3.11a and be can be written as: , (3.13a) . (3.13b) Note that neither Y’(+) nor Y’(-) spirals in the forbidden zone. And, since x is negative in the (left) forbidden zone, |Y’(+)| and |Y’(-)| both decay exponentially with increasing |x|. Summing Eqs. 3.13a and b produces: . (3.14) Therefore for x<0: . (3.15a) Differentiation produces: . (3.15b) At x=0: , (3.16a) . (3.16b) Comparison of Eqs. 3.7a and 3.16a shows that P is indeed continuous at x=0. And (see Eqs. 3.7b and 3.16b), the continuity of dP/dx at x=0 requires that . (3.17) Or, . (3.18) Now by definition, , (3.19) and thus . (3.20) Therefore . (3.21) Eq. 3.18 can therefore be rewritten as: . (3.22) Finally, . (3.23) As expected, in the limit of infinite Uo, f=0. Like an infinite barrier, a finite barrier will reflect all particles with kinetic energies less than the barrier height. Or, by deBroglie, all wave functions of particles meeting this condition will be reflected. However, in the finite barrier case the wave function penetrates into the forbidden zone. It is partially reflected in each interval (x<0,dx). Consequently the phase difference between Y’(+) and Y’(-) at the barrier will be p-2f … somewhat less than the value p found in the infinite barrier case. 4.  THE FINITE SQUARE WELL. Let us now consider the finite square well (see Fig. 2.1). It will be instructive to consider the first few energy states individually. We expect something less than half a wavelength to be contained in the Ground State (E=Eo). In the first excited state (E1) we expect something less than a full wavelength to be contained, etc. The Ground State Fig. 4.1 qualitatively depicts values for Y(+) and Y(-) at the well walls. Figure 4.1 Y(+) and Y(-), Ground State   The requirement that Y(+)(0) be unique suggests that Y(+)(0) must spiral (left-handed spiral) to a value of Y(+)(0)ei(p-f) at x=L (see Fig. 4.1). Similarly Y(-)(0)= Y(+)(0)ei(p-f) must spiral (right-handed spiral) to a value of Y(+)(0) at x=L. On reflection Y(+) experiences a phase change into Y(-) and Y(-) changes into Y(+). In a traversal of Y(+)’s spiral from x=0 to x=L, Y(+) rotates counterclockwise by two consecutive partial quarter turns. Thus in the Ground State less than a full half wave is contained between the finite well’s walls: . (4.1) Or, . (4.2) But . (4.3) Thus . (4.4) This transcendental equation in Eo is readily solved (using a computer) by "creeping backward" from the infinite well Ground State energy of h2/8mL2. The First Excited State. Fig. 4.2 qualitatively depicts suggested values for Y(+) and Y(-) at x=0 and x=L. In this case Y(+) rotates a partial quarter turn, two full quarter turns, and another partial quarter turn in going from x=0 to x=L. As usual, Y(+) and Y(-) reflect into each other at the well walls. Figure 4.2 First Excited State The angle f depends only on Uo and E1 (see Eq. 3.23), and Eq. 4.1 now becomes , (4.5) or . (4.6) In this case , (4.7) whence . (4.8) Again we can "creep backward" from the infinite well energy of h2/2mL2 in solving this equation. Second Excited State. Fig. 4.1 also depicts suggested values of Y(+) and Y(-) a , (4.9) and . (4.10) Thus . (4.11) Arbitrary En. For arbitrary En Em, etc. The good news is that quantum theory provides guidelines on what the probability of finding a system at one energy or another is. In general any given system will be constantly exchanging energy levels with its neighbors. All told there are many lines of inquiry where the spinning spiral model might lead to better understanding and even to new insights. The complex algebra/calculus of wave mechanics may at times seem daunting. Yet practically anyone can visualize a rotating spiral in his/her mind’s eye. Comments?  mailto:noxid100@cox.net